Re: [neonixie-l] Rookie question about driving VFD filament
Hello Ten > Another idea is to connect 4 tubes in series. Of course it will make a > voltage gradient across tubes, but I plan to drive anodes at 50+V, so maybe > it won't be visible - does anybody have any experience with driving > filaments like that? > A step-down converter sounds good, but I will be very space limited in > this project. > I cannot answer all your questions. In my 4x IV-6 clock I connected filaments in series, drove them DC and let the dropping resistor on the ground side of the series. This way the filament of "last" IV-6 is about 1V above ground so the condition for "totally off" is satisfied. So: +5V fil1 fil2 fil3 fil4 --- R GND On my todo list there's a 6x or 8x display with IV-3 or IV-6. A PCB is work in progress and I would like to insert extra pads to allow maximum flexibility on filament driving: AC, DC, series, series/parallel. HTH, Paolo -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+unsubscr...@googlegroups.com. To post to this group, send an email to neonixie-l@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/CABj2VaZ%2BHfjKGXaCeoGRBP1LPUc_6mpT3S_VvGbLdZ6z6%2B0hOw%40mail.gmail.com. For more options, visit https://groups.google.com/d/optout.
Re: [neonixie-l] Rookie question about driving VFD filament
> Do I understand correctly, that they work just like a resistor? It *is* a resistor, one with a largish positive temperature coefficient. > My plan is to drive IV-6 filament by directly PWMing it from 5V or 3,3V. It > normally requires 50mA@1V, so if the filament works just like a resistor, can > I PWM it with low duty cycle (4% on 5V and 9,2% on 3,3V)? Of course PWM > frequency will be something above 25kHz to avoid any noise. Will the wire > burn from short current spikes? > I don't like the idea of adding a dropper resistor. on 5V I will be losing 4x > the power needed to warm the cathodes! I prefer to drive them with AC. The idea is similar to PWM, but with two out of phase waveforms. I use a series capacitor to limit the current without wasting energy. A MAX628 or similar MOSFET driver makes a dandy driver, it has two channels, one inverting and one non-inverting, so a single PWM signal can provide both phases. It has plenty of current capability, so it can drive all the filaments in parallel (one series capacitor each to set/limit the current). A side advantage of this approach is that if you're driving it with (say) 5V, the average voltage is half that, so the filament is effectively at a small positive voltage. Then when you ground a grid to turn off a segment, it appears as a negative bias and turns it off solidly. - John -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+unsubscr...@googlegroups.com. To post to this group, send an email to neonixie-l@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/C3E013EC-FB71-4D12-B219-38290F4A3F6E%40mac.com. For more options, visit https://groups.google.com/d/optout.
Re: [neonixie-l] Rookie question about driving VFD filament
No, the filaments are tungsten and tungsten has a huge positive temperature coefficient. The ideal drive would be a constant current, voltage capped source. A resistor approximates that quite well. I would not connect them in series for the same reason it's a bad idea to connect vacuum tube filaments in series. I have an art deco tube radio that still works that has the tubes in series. Upon turn-on, one filament goes incandescent and then dims down to normal as the other tubes warm up. Needless to say, that tube doesn't last very long. John On 03/04/2017 03:34 AM, ten kowal wrote: > Do I understand correctly, that they work just like a resistor? > My plan is to drive IV-6 filament by directly PWMing it from 5V or 3,3V. It > normally requires 50mA@1V, so if the filament works just like a resistor, > can I PWM it with low duty cycle (4% on 5V and 9,2% on 3,3V)? Of course PWM > frequency will be something above 25kHz to avoid any noise. Will the wire > burn from short current spikes? > I don't like the idea of adding a dropper resistor. on 5V I will be losing > 4x the power needed to warm the cathodes! > Another idea is to connect 4 tubes in series. Of course it will make a > voltage gradient across tubes, but I plan to drive anodes at 50+V, so maybe > it won't be visible - does anybody have any experience with driving > filaments like that? > A step-down converter sounds good, but I will be very space limited in this > project. > -- John DeArmond Tellico Plains, Occupied TN http://www.tnduction.com<-- THE source for induction heaters http://www.neon-john.com<-- email from here http://www.johndearmond.com <-- Best damned Blog on the net PGP key: wwwkeys.pgp.net: BCB68D77 -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+unsubscr...@googlegroups.com. To post to this group, send an email to neonixie-l@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/d492bf1a-b1a6-e6cc-867f-cdb686033cb3%40neon-john.com. For more options, visit https://groups.google.com/d/optout.
Re: [neonixie-l] Rookie question about driving VFD filament
An important aspect of the filament has not been discussed in this thread: Yes, the filament needs to be heated so that it emits electrons efficiently but it also sets the potential difference between the cathode (Itself) and the anodes (the segments). For the later reason you can't simply calculate the number of watts and then use a PWM to get to that number. 20% of 5V is technically 1V but that means that one end of the cathode is at 5V potential (for 25% of the time) and the other end is at 0V. If the anodes are at 30V then the end of the display nearest the PWM will get 28V (anode to cathode) and the end nearest the ground side will get 30V (anode to cathode) or 7% brighter. This is why filaments, especially for wide multi digit displays, are operated with an AC drive to minimize the end-to-end voltage differential. For a single digit like the IV-6 this will result in uneven illumination between the left and right side of the display. Another consideration is the grid to filament bias. The grid for an "Off" digit is typically held at a negative potential relative to the filament to repel the emitted electrons thus preventing them from hitting the anodes. A grid that is not properly biased this way results in segments that are dimly lit when they should be off because of electrons that sneak through the grid. This is why the filament is usually biased so that ground is below the lowest voltage of either end of the filament so that grids at ground potential are negative with respect to the filament... or else the grids are operated at a negative voltage for filaments at zero volts. An IV-6 operated by itself does not need the grid to be used but for a group of them in a multiplexed display this is critical. If 5 filaments are connected in series to 5V, they will all get 1V but the cathode to anode potential (For a 30V anode voltage) for the one nearest the 5V will be 25.5V (Average) and the one nearest ground will be 0.5V (Average) or a 20% difference in brightness. -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+unsubscr...@googlegroups.com. To post to this group, send an email to neonixie-l@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/a9bf4901-b949-4f1c-8a90-7d477a85fc69%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.
Re: [neonixie-l] Rookie question about driving VFD filament
While on the subject of VFD's, anyone know of a particular VFD that has a long lifespan ? I'm reluctant to build a clock with them unless I'm certain they have a long lifetime, as I've seen the VFD on my kitchen stove is fading near the filaments (I suspect electron bombardment on the cathodes has caused phosphor damage, and it's at it's worst nearest the filament) -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+unsubscr...@googlegroups.com. To post to this group, send an email to neonixie-l@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/63b996f7-2f91-4fef-8ab9-75ccb787d44c%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.
RE: [neonixie-l] Rookie question about driving VFD filament
<20% of 5V is technically 1V > Please let me add a word of caution on the pwm issue here. Although the statement seems intuitively true, it falls apart when current (and therefore power) is taken in consideration. In order to calculate the required pulse width ratio the following formula must be used: Using the following definitions: d = t/T (ratio = on time / period time), R = filament resistance V = Voltage I = current P = power (I*V) Vdc = rated voltage of the filament Vpwm = supply Voltage First statement: Pdc = Ppwm ==> Vdc*Idc = d*Vpwm* ==>Vdc*(Vdc/R) = Vpwm*(Vpwm/R) cancel out R ==>Vdc^2 = d*Vpwm^2 than solving for d: d = (Vdc)^2 / (Vpwm)^2 S In order to drive a filament rated for 1V dc from a 5V dc supply d will equal (1^2/5^2)*100% = 4%. A 20% duty cycle WILL burn the filament in very short order. Bill From: neonixie-l@googlegroups.com [mailto:neonixie-l@googlegroups.com] On Behalf Of taylorjpt Sent: Tuesday, March 07, 2017 12:30 PM To: neonixie-l Subject: Re: [neonixie-l] Rookie question about driving VFD filament An important aspect of the filament has not been discussed in this thread: Yes, the filament needs to be heated so that it emits electrons efficiently but it also sets the potential difference between the cathode (Itself) and the anodes (the segments). For the later reason you can't simply calculate the number of watts and then use a PWM to get to that number. 20% of 5V is technically 1V but that means that one end of the cathode is at 5V potential (for 25% of the time) and the other end is at 0V. If the anodes are at 30V then the end of the display nearest the PWM will get 28V (anode to cathode) and the end nearest the ground side will get 30V (anode to cathode) or 7% brighter. This is why filaments, especially for wide multi digit displays, are operated with an AC drive to minimize the end-to-end voltage differential. For a single digit like the IV-6 this will result in uneven illumination between the left and right side of the display. Another consideration is the grid to filament bias. The grid for an "Off" digit is typically held at a negative potential relative to the filament to repel the emitted electrons thus preventing them from hitting the anodes. A grid that is not properly biased this way results in segments that are dimly lit when they should be off because of electrons that sneak through the grid. This is why the filament is usually biased so that ground is below the lowest voltage of either end of the filament so that grids at ground potential are negative with respect to the filament... or else the grids are operated at a negative voltage for filaments at zero volts. An IV-6 operated by itself does not need the grid to be used but for a group of them in a multiplexed display this is critical. If 5 filaments are connected in series to 5V, they will all get 1V but the cathode to anode potential (For a 30V anode voltage) for the one nearest the 5V will be 25.5V (Average) and the one nearest ground will be 0.5V (Average) or a 20% difference in brightness. -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+unsubscr...@googlegroups.com. To post to this group, send email to neonixie-l@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/a9bf4901-b949-4f1c-8a90-7d477a85fc69%40googlegroups.com <https://groups.google.com/d/msgid/neonixie-l/a9bf4901-b949-4f1c-8a90-7d477a85fc69%40googlegroups.com?utm_medium=email&utm_source=footer> . For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+unsubscr...@googlegroups.com. To post to this group, send an email to neonixie-l@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/000e01d2977a%248d3ff930%24a7bfeb90%24%40com. For more options, visit https://groups.google.com/d/optout.
RE: [neonixie-l] Rookie question about driving VFD filament
Oops, I did not complete one line: First statement: Pdc = Ppwm ==> Vdc*Idc = d*Vpwm*Ipwm <--<< ==>Vdc*(Vdc/R) = Vpwm*(Vpwm/R) cancel out R ==>Vdc^2 = d*Vpwm^2 than solving for d: d = (Vdc)^2 / (Vpwm)^2 From: neonixie-l@googlegroups.com [mailto:neonixie-l@googlegroups.com] On Behalf Of Bill van Dijk Sent: Tuesday, March 07, 2017 2:40 PM To: neonixie-l@googlegroups.com Subject: RE: [neonixie-l] Rookie question about driving VFD filament <20% of 5V is technically 1V > Please let me add a word of caution on the pwm issue here. Although the statement seems intuitively true, it falls apart when current (and therefore power) is taken in consideration. In order to calculate the required pulse width ratio the following formula must be used: Using the following definitions: d = t/T (ratio = on time / period time), R = filament resistance V = Voltage I = current P = power (I*V) Vdc = rated voltage of the filament Vpwm = supply Voltage First statement: Pdc = Ppwm ==> Vdc*Idc = d*Vpwm* ==>Vdc*(Vdc/R) = Vpwm*(Vpwm/R) cancel out R ==>Vdc^2 = d*Vpwm^2 than solving for d: d = (Vdc)^2 / (Vpwm)^2 S In order to drive a filament rated for 1V dc from a 5V dc supply d will equal (1^2/5^2)*100% = 4%. A 20% duty cycle WILL burn the filament in very short order. Bill From: neonixie-l@googlegroups.com [mailto:neonixie-l@googlegroups.com] On Behalf Of taylorjpt Sent: Tuesday, March 07, 2017 12:30 PM To: neonixie-l Subject: Re: [neonixie-l] Rookie question about driving VFD filament An important aspect of the filament has not been discussed in this thread: Yes, the filament needs to be heated so that it emits electrons efficiently but it also sets the potential difference between the cathode (Itself) and the anodes (the segments). For the later reason you can't simply calculate the number of watts and then use a PWM to get to that number. 20% of 5V is technically 1V but that means that one end of the cathode is at 5V potential (for 25% of the time) and the other end is at 0V. If the anodes are at 30V then the end of the display nearest the PWM will get 28V (anode to cathode) and the end nearest the ground side will get 30V (anode to cathode) or 7% brighter. This is why filaments, especially for wide multi digit displays, are operated with an AC drive to minimize the end-to-end voltage differential. For a single digit like the IV-6 this will result in uneven illumination between the left and right side of the display. Another consideration is the grid to filament bias. The grid for an "Off" digit is typically held at a negative potential relative to the filament to repel the emitted electrons thus preventing them from hitting the anodes. A grid that is not properly biased this way results in segments that are dimly lit when they should be off because of electrons that sneak through the grid. This is why the filament is usually biased so that ground is below the lowest voltage of either end of the filament so that grids at ground potential are negative with respect to the filament... or else the grids are operated at a negative voltage for filaments at zero volts. An IV-6 operated by itself does not need the grid to be used but for a group of them in a multiplexed display this is critical. If 5 filaments are connected in series to 5V, they will all get 1V but the cathode to anode potential (For a 30V anode voltage) for the one nearest the 5V will be 25.5V (Average) and the one nearest ground will be 0.5V (Average) or a 20% difference in brightness. -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+unsubscr...@googlegroups.com. To post to this group, send email to neonixie-l@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/a9bf4901-b949-4f1c-8a90-7d477a85fc69%40googlegroups.com <https://groups.google.com/d/msgid/neonixie-l/a9bf4901-b949-4f1c-8a90-7d477a85fc69%40googlegroups.com?utm_medium=email&utm_source=footer> . For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+unsubscr...@googlegroups.com. To post to this group, send email to neonixie-l@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/000e01d2977a%248d3ff930%24a7bfeb90%24%40com <https://groups.google.com/d/msgid/neonixie-l/000e01d2977a%248d3ff930%24a7bfeb90%24%40com?utm_medium=email&utm_source=footer> . For more options, visit https://groups.google.com/d/optout. -- You received this message because you
Re: [neonixie-l] Rookie question about driving VFD filament
Thanks everyone for answers, I will probably use some cheap VFD to test the 9.2% PWM @3,3V on high frequency. I'll measure current and will look on it on oscilloscope to see on which frequency the filament will be heated so frequently, that it won't have time to really cool (current will be flat enough on all ON time). I will also test if any kind of soft-start is needed (rushing 3,3V through cold filament sounds risky). as for the aspect of uneven lightning while PWMing - as far as I understand it doesn't change much if I PWM it or not, of course filament-anode voltage will be different, but on such a small tube with 50+ anode voltage I doubt it will make a change. Also, it definitely won't be visible as left-right difference, because in IV-6 filament is a single wire going directly in the middle from top to bottom - so depending on connection I can make top or bottom a bit dimmer. I've found one project, where author built a H bridge to be able to imitate AC. I think that it might be the way to go, combining H bridging with PWM. Definitely gonna try it some day - I must find a lot of cheap VFDs, here in my country there is someone offering 100x IV-8 for 30 dollars, I'll probably just buy them to have enough "material". https://callumnunesvaz.wordpress.com/portfolio/dogbonevfd/ W dniu wtorek, 7 marca 2017 18:30:16 UTC+1 użytkownik taylorjpt napisał: > > An important aspect of the filament has not been discussed in this > thread: Yes, the filament needs to be heated so that it emits electrons > efficiently but it also sets the potential difference between the cathode > (Itself) and the anodes (the segments). For the later reason you can't > simply calculate the number of watts and then use a PWM to get to that > number. 20% of 5V is technically 1V but that means that one end of the > cathode is at 5V potential (for 25% of the time) and the other end is at > 0V. If the anodes are at 30V then the end of the display nearest the PWM > will get 28V (anode to cathode) and the end nearest the ground side will > get 30V (anode to cathode) or 7% brighter. This is why filaments, > especially for wide multi digit displays, are operated with an AC drive to > minimize the end-to-end voltage differential. For a single digit like the > IV-6 this will result in uneven illumination between the left and right > side of the display. > > Another consideration is the grid to filament bias. The grid for an "Off" > digit is typically held at a negative potential relative to the filament to > repel the emitted electrons thus preventing them from hitting the anodes. > A grid that is not properly biased this way results in segments that are > dimly lit when they should be off because of electrons that sneak through > the grid. This is why the filament is usually biased so that ground is > below the lowest voltage of either end of the filament so that grids at > ground potential are negative with respect to the filament... or else the > grids are operated at a negative voltage for filaments at zero volts. An > IV-6 operated by itself does not need the grid to be used but for a group > of them in a multiplexed display this is critical. If 5 filaments are > connected in series to 5V, they will all get 1V but the cathode to anode > potential (For a 30V anode voltage) for the one nearest the 5V will be > 25.5V (Average) and the one nearest ground will be 0.5V (Average) or a 20% > difference in brightness. > > > > > -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+unsubscr...@googlegroups.com. To post to this group, send an email to neonixie-l@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/8c816c3c-10b8-4117-963e-a89a1be7bf29%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.
Re: [neonixie-l] Rookie question about driving VFD filament
> Thanks everyone for answers, I will probably use some cheap VFD to test the > 9.2% PWM @3,3V on high frequency. I'll measure current and will look on it on > oscilloscope to see on which frequency the filament will be heated so > frequently, that it won't have time to really cool (current will be flat > enough on all ON time). I will also test if any kind of soft-start is needed > (rushing 3,3V through cold filament sounds risky). This is where a series resistor or capacitor comes in handy. They will limit the amount of current allowed to flow through a cold filament. > I've found one project, where author built a H bridge to be able to imitate > AC. My suggestion of using a MOSFET driver amounts to the same thing (it's not original to me, I lifted the idea from Konstantin). > I think that it might be the way to go, combining H bridging with PWM. I'm fond of it. If you use a capacitor as the dropping element, current goes to zero if PWM is off, which is an easy way to turn off the filament. - John -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+unsubscr...@googlegroups.com. To post to this group, send an email to neonixie-l@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/EF36EC5D-4EC4-4127-9437-A6CB0CC36534%40mac.com. For more options, visit https://groups.google.com/d/optout.